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title: 'Math 181 Miniproject 2: Population and Dosage'

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Math 181 Miniproject 2: Population and Dosage.md
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Math 181 Miniproject 2: Population and Dosage
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**Overview:**  In this miniproject you will use technological tools to turn data and into models of real-world quantitative phenomena, then apply the principles of the derivative to them to extract information about how the quantitative relationship changes.

**Prerequisites:** Sections 1.1--1.6 in *Active Calculus*, specifically the concept of the derivative and how to construct estimates of the derivative using forward, backward and central differences. Also basic knowledge of how to use Desmos. 

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1\. A settlement starts out with a population of 1000. Each year the population increases by $10\%$. Let $P(t)$ be the function that gives the population in the settlement after $t$ years.
(a) Find the missing values in the table below.
:::

(a)
| $t$    | 0  | 1  | 2  | 3  | 4  | 5  | 6  | 7  |
|--------|----|----|----|----|----|----|----|----|
| $P(t)$ |1000|1100|1210|1331|1464|1611|1772|1949|





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(b) Find a formula for $P(t)$. You can reason it out directly or you can have Desmos find it for you by creating the table of values above (using $x_1$ and $y_1$ as the column labels) and noting that the exponential growth of the data should be modeled using an exponential model of the form
\\[
y_1\sim a\cdot b^{x_1}+c
\\]
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(b)
\\[y_1\sim 999.767\cdot 1.10006^{x_1}+0.182344\\] 



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(c\) What will the population be after 100 years under this model?
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(c\)
\\[y_1\sim 999.767\cdot 1.10006^{100}+0.182344\\]
\\[y_1\sim 999.767\cdot 13855.98263+0.182344\\]
$y=13852754.37 people$
 



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(d) Use a central difference to estimate the values of $P'(t)$ in the table below. What is the interpretation of the value $P'(5)$?
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(d)
$P'(5)=\frac{P(6)-P(4)}{6-4}=\frac{1772-1464}{2}=154$ people/year
| $t$     | 1 |  2  |  3  |  4  |  5  |  6  |
|---      |---|-----|-----|-----|-----|-----|
| $P'(t)$ |105|115.5| 127 |139.5| 154 |169  |
My interpretaion of P'(5) is at year 5, the population has increased by 154 people, which shows that the population is gradually increasing.



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(e) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value?
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(e)
$P''(3)=\frac{P(4)-P(2)}{4-2}=\frac{139.5-115.5}{2}=12$ $poeple/year^2$
My interpretation of this value is that the population is increasing and at year 3, the population has increased by a rate of 12 $people per  year^2$
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(f) **Cool Fact:** There is a constant $k$ such that $P'(t)=k\cdot P(t)$. In other words, $P$ and $P'$ are multiples of each other.

What is the value of $k$? (You could try creating a slider and playing with the graphs or you can try an algebraic approach.)
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(f)
$P'(t)=k\cdot P(t)$
$P'(105)=k\cdot P(1100)$
divide each side by 1100
k= 0.0954545455
You can double check this by multiplying the calculated value of k by P(1100)
$0.0954545455\cdot 1100 =105 people/year$




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2\. The dosage recommendations for a certain drug are based on weight.
| Weight (lbs)| 20 | 40 | 60 | 80  | 100 | 120 | 140 | 160 | 180 |
|---          |--- |--- |--- |---  |---  |---  |---  |---  |--- |
| Dosage (mg) | 10 | 30 | 70 | 130 | 210 | 310 | 430 | 570 | 730 |
(a) Find a function D(x) that approximates the dosage when you input the weight of the individual. (Make a table in Desmos using $x_1$ and $y_1$ as the column labels and you will see that the points seem to form a parabola. Use Desmos to find a model of the form
\\[
y_1\sim ax_1^2+bx_1+c
\\]
and define $D(x)=ax^2+bx+c$.)
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(a)
\\[
y_1\sim 0.025x_1^2-0.5x_1+10
\\]
a stands for the type of graph, which in this case is a parabola and it also shows that the graph will open wider with a smaller slope than the parent function would have been 
b could be defined as,
c stands for the y intercept of the graph




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(b) Find the proper dosage for a 128 lb individual.
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(b)
\\[
y_1\sim 0.025(128)^2-0.5(128)+10
\\] $=355.6 mg$
 




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(c\) What is the interpretation of the value $D'(128)$. 
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(c\)
My interpretation of D'(128) is
D'(x)= 0.05x-0.5
D'(128)= 0.05(128)-0.5
D'(128)= 5.9mg/lb
Beyond the weight of 128lbs a persons dosage would increase at the rate of 5.9mg/lb
I got the above equation by calculating the limit definition of a derivative which gave me the above previous answer 
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(d) Estimate the value of $D'(128)$ using viable techniques from our calculus class. Be sure to explain how you came up with your estimate.
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(d)
$In part c. I kind of answered this question as well. Above I used the limit defintion of a derivative to calculate an approximate of 128lbs. Above I showed my work and got an answer of 5.9mg/lb.
This approximation means that starting at the weight of 128lbs the approximate rate of change of dosage is going to be 5.9mg/lb. To be more exact, I got the answer of 5.9mg/lb but using the equation found in part a to put in the limit definition of a derivative formula. After doing the algebra for that I got 0.05x-.5 and plugged in x for 128 to get 5.9 mg/lb


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(e) Given the value $D'(130)=6$, find an equation of the tangent line to the curve $y=D(x)$ at the point where $x=130$ lbs. 
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(e)
$D(130)= 0.025(130)^2-0.5(130)+10$
D(130)=367.5mg

$y-m(x-x^1)$
y=367.5+6(x-130)
y=367.5+6x-780
$6x-y+412.5$ this is the equation to the tangent line
Or it could be written as $y=6x+412.5$ 
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(f) Find the point on the tangent line in the previous part that has $x$-coordinate $x=128$. Does the output value on the tangent line for $x=128$ lbs give a good estimate for the dosage for a 128 lb individual? 
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(f)
To solve this plug in x=128 into the previous equation that we solved for to find a good estimate
y=6x-412.5
y= 6(128) -412.5
y= 355.5 mg
This is the estimate for dosage at the weight of 128 lbs, and yes it is a good estimate because it's very close to the actual value. 



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